J. Nonlinear Sci. Appl. 4 (2011), no. 3, 193–199
The Journal of Nonlinear Sciences and Applications
http://www.tjnsa.com
INTERNAL NONLOCAL AND INTEGRAL CONDITION
′
′
PROBLEMS OF THE DIFFERENTIAL EQUATION x = f (t, x, x )
A. M. A. EL-SAYED1 , E. M. HAMDALLAH2 AND KH. W. ELKADEKY3
Abstract. In this work, we are concerned with the existence of at least
one absolutely continuous solution of the Cauchy problem for the differen′
′
tial
∑m equation x = f (t, x, x ), t ∈ (0, 1) with the internal nonlocal condition
) = xo , τk ∈ (c, d) ⊆ (0, 1). The problem of the integral
k=1 ak x(τ
∫ dk
condition c x(s) dg(s) = xo will be considered.
1. Introduction
Problems with non-local conditions have been extensively studied by several
authors in the last two decades. The reader is referred to [1]- [6] and [9] - [15]
and references therein.
Here we are consisted with the nonlocal problem
dx(t)
dx(t)
= f (t, x(t),
), a.e, t ∈ (0, 1),
dt
dt
m
∑
k=1
ak x(τk ) = xo ,
m
∑
ak ̸= 0 and τk ∈ (c, d) ⊆ (0, 1).
(1.1)
(1.2)
k=1
The existence of at least one solution x ∈ AC[0, 1] will be studied when the
function f is measurable in t ∈ [0, 1],for any (u1 , u2 ) ∈ R2 and continuous in
(u1 , u2 ) ∈ R2 , for t ∈ [0, 1]. As a consequence of our result, the problem of the
Date: Received: 1 Jan, 2011; Revised: 20 April, 2011.
∗
Corresponding author
c 2011 N.A.G.
⃝
2000 Mathematics Subject Classification. Primary 26A33; Secondary 34A12, 34A40.
Key words and phrases. Nonlocal conditions, integral condition, existence of solution, fixedpoint theorem.
193
194
A. M. A. EL-SAYED, E. M. HAMDALLAH AND KH. W. ELKADEKY
differential equation (1.1) with integral condition
∫ d
x(s) dg(s) = xo ,
(1.3)
c
where g is a nondecreasing function, will be studied.
It must be noticed that the nonlocal condition (1.2) and the integral condition
(1.3) are more general than the following ones
x(τ ) = xo , τ ∈ (c, d),
m
∑
(1.4)
ak x(τk ) = 0, τk ∈ (a, c),
(1.5)
k=1
∫
and
c
x(s) dg(s) = 0.
(1.6)
a
The following theorems will be needed.
Theorem (Kolmogorov Compactness Criterion) see[8]
Let Ω ⊆ LP (0, 1), 1 ≤ P < ∞. If
(i) Ω is bounded Lp (0, 1),
(ii) xh → x as h → 0 uniformly with respect to x ∈ Ω, then Ω is relatively
compact in LP (0, 1), where
∫
1 t+h
xh (t) =
x(s) ds.
h t
Theorem (Schauder) see[12]
Let U be a convex subset of a Banach space X, and T : U → U is compact,
continuous map. Then T has at least one fixed point in U .
2. Existence of solution
The following Lemma gives the integral equation representation for the nonlocal problem (1.1)-(1.2).
Lemma 2.1 The solution of the nonlocal problem (1.1)-(1.2) can be expressed
by the integral equation
∫ τk
∫ t
m
∑
x(t) = ax0 − a
ak
y(s) ds +
y(s) ds
(2.1)
k=1
0
0
where y is the solution of the functional integral equation
∫ τk
∫ t
m
∑
y(t) = f (t, ax0 − a
ak
y(s) ds +
y(s) ds, y(t)), t ∈ (0, 1).
k=1
Proof. Let
dx(t)
dt
0
(2.2)
0
= y(t) in equation (1), then
∫ t
y(t) = f (t, x(0) +
y(s) ds, y(t))
0
(2.3)
INTERNAL NONLOCAL AND INTEGRAL CONDITION PROBLEMS
where
∫
195
t
x(t) = x(0) +
y(s) ds.
(2.4)
0
Let t = τk in (2.4), we obtain
∫
τk
x(τk ) = x(0) +
y(s) ds
0
and
m
∑
ak x(τk ) =
k=1
m
∑
ak x(0) +
k=1
∫
m
∑
ak
τk
y(s) ds.
(2.5)
0
k=1
Substitute from (2) into (2.5), we get
x0 =
m
∑
ak x(0) +
k=1
m
∑
∫
ak
y(s) ds
0
k=1
and
x(0) = a(x0 −
m
∑
k=1
τk
∫
τk
ak
y(s) ds)
(2.6)
0
∑
−1
where a = ( m
.
k=1 ak )
Substitute from (2.6) into (2.4) and (2.3), we obtain (2.1) and (2.2).
Consider the functional integral equation (2.2) with the following assumptions
(i) f : [0, 1] × R2 → R+ is measurable in t ∈ [0, 1] for any (u1 , u2 ) ∈ R2
and continuous in (u1 , u2 ) ∈ R2 for almost all t ∈ [0, 1].
(ii) There exists a function a ∈ L1 [0, 1] and two constants bi > 0 , i = 1, 2
such that
|f (t, u1 , u2 )| ≤ |a(t)| +
2
∑
bi |ui |, ∀(t, u1 , u2 ) ∈ [0, 1] × R2 .
i=1
(iii)
(2b1 + b2 ) < 1.
Now we have the following theorem.
Theorem 2.1 Assume that the assumptions (i) - (iii) are satisfied. Then the
functional integral equation (2.2) has at least one solution y ∈ L1 (0, 1).
||a||+ab1 x0
Proof. Let y ∈ Br ⊂ L1 , Br = {y : ||y||L1 ≤ r, r > 0}, r = 1−(2b
.
1 +b2 )
Clearly Br is nonempty, convex and closed .
Define the operator H by
∫ τk
∫ t
m
∑
y(s) ds +
y(s) ds, y(t)), t ∈ (0, 1). (2.7)
Hy(t) = f (t, ax0 − a
ak
k=1
0
0
196
A. M. A. EL-SAYED, E. M. HAMDALLAH AND KH. W. ELKADEKY
From assumptions (i) and (iii), we obtain
∫ 1
||Hy||L1 =
|(Hy)(t)| dt
0
∫
1
|f (t, ax0 − a
=
0
∫
m
∑
∫
ak
k=1
1
≤
(|a(t)| + b1 |axo − a
0
∫
≤
∫
1
∫
0
∫
m
∑
0
∫
t
|y(s)| dsdt + b2
+ b1
0
0
∫
τk
m
∑
∫
1
0
∫
τk
ak
k=1
0
t
y(s) ds| + b2 |y(t)|) dt
y(s) ds +
b1 axo dt + ab1
0
y(s) ds, y(t))| dt
0
ak
1
(|a(t)|dt +
0
t
y(s) ds +
k=1
∫
1
∫
τk
|y(s)| dsdt
0
1
|y(t)|) dt
o
≤ ||a|| + ab1 xo + b1 ||y|| + b1 ||y|| + b2 ||y||
≤ ||a|| + b1 ax0 + (2b1 + b2 )||y|| ≤ r.
Then ||Hy||L1 ≤ r, which implies that the operator H maps Br into itself.
Assumption (i) implies that H is continuous.
Now, let Ω be a bounded subset of Br , therefore H(Ω) is bounded in L1 (0, 1),
i.e condition (i) of Kolmogorav compactness criterion is satisfied, it remains to
show (Hy)h → (Hy), in L1 (0, 1].
Let y ∈ Ω ⊂ L1 (0, 1), then we have the following
∫ 1
||(Hy)h − (Hy)||L1 =
|(Hy)h (t) − (Hy)(t)| dt
0
∫ 1 ∫ t+h
1
(Hy)(s) ds − (Hy)(t)| dt
=
|
h t
0
∫
∫ 1
1 t+h
|(Hy)(s) − (Hy)(t)| ds) dt
≤
(
h t
0
∫ 1 ∫ t+h
∫ sk
∫ s
m
∑
1
≤
|f (s, ax0 − a
ak
y(τ )dτ +
y(τ )dτ, y(s))
h t
0
0
0
k=1
∫ τk
∫ t
m
∑
− f (t, ax0 − a
ak
y(s)ds +
y(s)ds, y(t))| ds dt.
k=1
0
0
Since y ∈ Ω ⊂ L1 , and (assumption (ii) implies that) f ∈ L1 [0, 1], it follows
that
∫ sk
∫ s
∫
m
∑
1 t+h
y(τ )dτ +
y(τ )dτ, y(s))
|f (s, ax0 − a
ak
h t
0
0
k=1
INTERNAL NONLOCAL AND INTEGRAL CONDITION PROBLEMS
−f (t, ax0 − a
m
∑
∫
ak
k=1
∫
τk
t
y(s)ds, y(t))| ds → 0 as h → 0, t ∈ (0, 1).
y(s)ds +
0
197
0
Hence (Hy)h → (Hy), uniformly as h → 0.
Then by Kolmogorav compactness criterion, H(Ω) is relatively compact.
That is H has a fixed point in Br , then there exist at least one solution
y ∈ L1 (0, 1) of the functional equation (2.3).
Now, consider the nonlocal problem (1.1)-(1.2).
Theorem 2.2 Let the assumptions of Theorem 2.1 are satisfied. Then the
nonlocal problem (1.1)-(1.2) has at least one solution x ∈ AC[0, 1] .
Proof. Form Theorem 2.1 and equations (2.1) and (2.6) we deduce that there
exist at least one solution x ∈ AC[0, 1] of equation (2.1) where
∫ τk
m
∑
x(0) = lim x(t) = ax0 − a
ak
y(s) ds
(2.8)
t→0
0
k=1
and
x(1) = lim x(t) = ax0 − a
t→1
m
∑
∫
∫
τk
ak
y(s) ds +
y(s) ds
0
k=1
1
(2.9)
0
To complete the proof, we prove that equation (2.1) satisfies nonlocal problem
(1.1)-(1.2).
Differentiating (2.1), we get
dx
dx
= y(t) = f (t, x(t), )
dt
dt
Let t = τk in (2.1) ,we get
x(τk ) = ax0 − a
m
∑
∫
∫
τk
ak
0
k=1
= ax0 + (1 − a
τk
y(s) ds +
m
∑
k=1
y(s) ds
0
∫
ak )
τk
y(s) ds.
0
Then
m
∑
k=1
ak x(tk ) =
m
∑
k=1
ak ax0 +
m
∑
k=1
ak (1 − a
m
∑
k=1
∫
ak )
τk
y(s) ds = xo .
0
This complete the proof of the equivalent between the nonlocal problem (1.1)(1.2) and the integral equation (2.1) .
This implies that there exist at least one solution x ∈ AC[0, 1] of the nonlocal
problem (1.1)-(1.2).
198
A. M. A. EL-SAYED, E. M. HAMDALLAH AND KH. W. ELKADEKY
3. Nonlocal integral condition
Let x ∈ AC[0, 1] be the solution of the nonlocal problem (1.1)-(1.2). Let
ak = g(tk ) − g(tk−1 ), g is a nondecreasing function, τk ∈ (tk−1 , tk ), c = t0 <
t1 < t2 , ... < tn = d, then the nonlocal condition (1.2) will be
m
∑
(g(tk ) − g(tk−1 )) x(τk ) = xo .
k=1
From the continuity of the solution x of the nonlocal problem (1.1)-(1.2) we can
obtain
lim
m→∞
m
∑
∫
(g(tk ) − g(tk−1 )) x(τk ) =
d
x(s) dg(s).
c
k=1
and the nonlocal condition (2) transformed to the integral one
∫
d
x(s) dg(s) = xo .
c
Now, we have the following Theorem
Theorem 3.1 Let the assumptions of Theorem 2.1 are satisfied. Then there
exists at least one solution x ∈ AC[0, 1] of the nonlocal problem with integral
condition,
dx(t)
dx(t)
= f (t, x(t),
), a.e, t ∈ (0, 1],
dt
dt
∫ d
x(s) dg(s) = xo .
c
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1
Department of Mathematics, Alexandria University, Alexandria, Egypt
E-mail address: amasayed@hotmail.com
2
Department of Mathematics, Alexandria University, Alexandria, Egypt
E-mail address: emanhamdalla@hotmail.com
3
Department of Mathematics, Faculty of Science, Garyounis University, Benghazi, Libya
E-mail address: k-welkadeky@yahoo.com